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Anastasy [175]
3 years ago
15

Please help out asap! AWARDING 30 POINTS IF YOU TELL ME THE CORRECT ANSWER!

Mathematics
1 answer:
Anettt [7]3 years ago
5 0

Answer: Amy and Jeffery should choose the offer with the lowest interest rate and monthly payment under $1400. The answer would be Offer 2.

Explanation: All of the things are lower. Offer 3 is out because 1581.59. Offer 1 is out because it’s higher then offer 2.

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Find the area of the following shapes:
Annette [7]

Answer:

24 square units

Step-by-step explanation:

You apply area of a trapezium

5 0
3 years ago
In Mrs Phillips math class the ratio of boys to girls is 4 to 2. If there are 51 girls in math class, how many boys are there?
maksim [4K]
So for every two girls, there's four boys. There are twice as many boys than girls, so the number of girls • 2 = your answer.

51 • 2 = 102
7 0
3 years ago
Is -18x - 7 equal to -6(3x - 7)?
adoni [48]

Answer:

No

Step-by-step explanation:

-18-7 is not equal to -6(3x-8) beacuse

-6(3x-8)= -18x+42

its obivous that the two values are not same

5 0
3 years ago
Read 2 more answers
(Kinda need help ASAP)
denis23 [38]

Answer:

Plzzzz answer fast......

4 0
2 years ago
The amount of time a passenger waits at an airport check-in counter is random variable with mean 10 minutes and standard deviati
Stolb23 [73]

Answer:

(a) less than 10 minutes

= 0.5

(b) between 5 and 10 minutes

= 0.5

Step-by-step explanation:

We solve the above question using z score formula. We given a random number of samples, z score formula :

z-score is z = (x-μ)/ Standard error where

x is the raw score

μ is the population mean

Standard error : σ/√n

σ is the population standard deviation

n = number of samples

(a) less than 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Therefore, the probability that the average waiting time waiting in line for this sample is less than 10 minutes = 0.5

(b) between 5 and 10 minutes

i) For 5 minutes

x = 5 μ = 10, σ = 2 n = 50

z = 5 - 10/2/√50

z = -5 / 0.2828427125

= -17.67767

P-value from Z-Table:

P(x<5) = 0

Using the z table to find the probability

P(z ≤ 0) = P(z = -17.67767) = P(x = 5)

= 0

ii) For 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Hence, the probability that the average waiting time waiting in line for this sample is between 5 and 10 minutes is

P(x = 10) - P(x = 5)

= 0.5 - 0

= 0.5

3 0
3 years ago
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